To multiply the polynomials ( (9x^2 + 10x + 4) ) and ( (9x^2 + 5x + 1) ), you can use the distributive property (also known as the FOIL method for binomials). Multiply each term in the first polynomial by each term in the second polynomial, then combine like terms. The resulting polynomial will be a degree 4 polynomial. For the full expansion, the result is ( 81x^4 + 85x^3 + 49x^2 + 20x + 4 ).
5x + 4 = 10x - 5 10x - 5x = 4 + 5 5x = 9 x = 1.8
6-3x = 5x-10x+2 4 = 8x-10x 4 = -2x x = -2
10x - 18 = 5x + 2 Add -5x + 18 to both sides of the equation, to get: 5x = 20, from which x = 4.
10x + 2y + 8z - 5x + 4z - 4y = 10x - 5x + 2y - 4y + 8z + 4z = (10 - 5)x + (2 - 4)y + (8 + 4)z = 5x - 2y + 12z
(3x)^(2-4) = (3x)^-2 = (9x2)^-1 = 1/(9x2)
5x + 4 = 10x - 5 10x - 5x = 4 + 5 5x = 9 x = 1.8
6-3x = 5x-10x+2 4 = 8x-10x 4 = -2x x = -2
10x - 18 = 5x + 2 Add -5x + 18 to both sides of the equation, to get: 5x = 20, from which x = 4.
It is: 10x^2 -4x = 2x(5x-2)
10x + 2y + 8z - 5x + 4z - 4y = 10x - 5x + 2y - 4y + 8z + 4z = (10 - 5)x + (2 - 4)y + (8 + 4)z = 5x - 2y + 12z
10x + 4/8
6-3x = 5x-10x+10 6-3x = -5x+10 -3x+5x = 10-6 2x = 4 x = 2
(3x)^(2-4) = (3x)^-2 = (9x2)^-1 = 1/(9x2)
Well, you can factor this. (10X^2 + 5X)/5x 5X(2X + 1)/5X cancel 5X; top and bottom 2X + 1 set to 0 and see 2X+ 1 = 0 2X = - 1 X = -1/2 -----------check to see what we get [10(- 1/2)^2 + 5(- 1/2)/5(- 1/2) 10/4 - 5/2/- 5/2 5/2 - 5/2/- 5/2 = 0 ----------- I think so because my TI-84 agrees with this answer
5x/x-2
(5x - 2)(25x2 + 10x + 4)
(5x - 2)(25x^2 + 10x + 4)